Goldbach Conjecture (7th june 1742) We note P∗ the odd prime numbers set. P∗ = {p1 = 3, p2 = 5, p3 = 7, p4 = 11, . . .} ∀ n ∈ 2N\{0, 2, 4}, ∃ p ∈ P∗ , p ≤ n/2, ∃ q ∈ P∗ , q ≥ n/2, n =p+q We call n’s Goldbach decomposition such a sum p + q. p and q are said n’s Goldbach decomponents. verified by computer until 4.1018 (Oliveira e Silva, 4.4.2012) Denise Vella-Chemla
Goldbach Conjecture Study
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Notations In the following, C.G. signifies Goldbach Conjecture, s.c. signifies congruences system, p.a. signifies arithmetic progression, T.r.c. signifies Chinese Remainders Theorem. For a given n, we note : n P∗1 (n) = {x ∈ P∗ /x ≤ } 2 √ ∗ ∗ P2 (n) = {x ∈ P /x ≤ n}
Denise Vella-Chemla
Goldbach Conjecture Study
23/5/2012
2 / 31
Reformulation
Goldbach Conjecture is equivalent to the following statement : ∀ n ∈ 2N\{0, 2, 4}, ∃ p ∈ P∗1 (n), ∀ m ∈ P∗2 (n), p 6≡ n (mod m)
Indeed, ∀ n ∈ 2N\{0, 2, 4}, ∃ p ∈ P∗1 (n), ∀ m ∈ P∗2 (n), p 6≡ n (mod m) ⇔ n − p 6≡ 0 (mod m) ⇔ n − p is prime
Denise Vella-Chemla
Goldbach Conjecture Study
23/5/2012
3 / 31
Examples study : Example 1 Why 19 is the smallest 98’s Goldbach decomponent ? 98 ≡ 98 ≡ 98 ≡ 98 ≡ 98 ≡ 98 ≡
3 (mod 5) 5 (mod 3) 7 (mod 7) 11 (mod 3) 13 (mod 5) 17 (mod 3)
98 6≡ 19 (mod 3) 98 6≡ 19 (mod 5) 98 6≡ 19 (mod 7)
(98-3=95 and 5 | 95) (98-5=93 and 3 | 93) (98-7=91 and 7 | 91) (98-11=87 and 3 | 87) (98-13=85 and 5 | 85) (98-17=81 and 3 | 81)
(98-19=79 and 3 6 | 79) (98-19=79 and 5 6 | 79) (98-19=79 and 7 6 | 79)
Conclusion : ∀ m ∈ P∗2 (98), 19 6≡ 98 (mod m) 19 is a 98’s Goldbach decomponent. Indeed, 98 = 19 + 79 with 19 and 79 both primes. Denise Vella-Chemla
Goldbach Conjecture Study
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Examples study : Example 2 Why 3 is a 40’s Goldbach decomponent ? Z/3Z
0 1 2
Z/5Z
0 1 2 3 4
Z/7Z
0 1 2 3 4 5 6
Z/11Z 0 1 2 3 4 5 6 7 8 9 10 3’s equivalence class in each finite field, 40’s equivalence class in each finite field. Conclusion : ∀ m ∈ P∗2 (40), 3 6≡ 40 (mod m) 3 is a 40’s Goldbach decomponent. Indeed, 40 = 3 + 37 with 3 and 37 primes. Denise Vella-Chemla
Goldbach Conjecture Study
23/5/2012
5 / 31
Examples study : Example 3 We are looking for Goldbach decomponents of natural integers that are ≡ 2 (mod 3) and ≡ 3 (mod 5) and ≡ 3 (mod 7). Those numbers we are looking Goldbach decomponents for, are natural integers of the form 210k + 38 (result provided by Chinese Remainders Theorem as we will see it sooner). We saw that odd prime natural integers p that are 6≡ 2 (mod 3) and 6≡ 3 (mod 5) and 6≡ 3 (mod 7) can be Goldbach decomponents of those numbers. If we omit the case of “little prime numbers” (i.e. congruences cases to 0 modulo one odd prime and only one), • p must be ≡ 1 (mod 3). • p must be ≡ 1 or 2 or 4 (mod 5). • p must be ≡ 1 or 2 or 4 or 5 or 6 (mod 7). Denise Vella-Chemla
Goldbach Conjecture Study
23/5/2012
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Examples study : Example 3 We are looking for Goldbach decomponents of some even natural integers ≡ 2 (mod 3) and ≡ 3 (mod 5) and ≡ 3 (mod 7) (⇔ of the form 210k + 38)
Combining all differents possibilities, we obtain : 1 1 1 1 1 1 1 1 1 1 1 1 1 1
(mod (mod (mod (mod (mod (mod (mod (mod (mod (mod (mod (mod (mod (mod
3) 3) 3) 3) 3) 3) 3) 3) 3) 3) 3) 3) 3) 3)
1 1 1 1 1 2 2 2 2 2 4 4 4 4
(mod (mod (mod (mod (mod (mod (mod (mod (mod (mod (mod (mod (mod (mod
5) 5) 5) 5) 5) 5) 5) 5) 5) 5) 5) 5) 5) 5)
1 2 4 5 6 1 2 4 5 6 1 2 4 5
(mod (mod (mod (mod (mod (mod (mod (mod (mod (mod (mod (mod (mod (mod
7) 7) 7) 7) 7) 7) 7) 7) 7) 7) 7) 7) 7) 7)
1 (mod 3) 4 (mod 5) 6 (mod 7) Denise Vella-Chemla
Goldbach Conjecture Study
23/5/2012
7 / 31
Examples : Example 3 We are looking for Goldbach decomponents of some even natural integers ≡ 2 (mod 3) and ≡ 3 (mod 5) and ≡ 3 (mod 7) (⇔ of the form 210k + 38)
Combining all differents possibilities, we obtain : 7) 7) 7) 7) 7) 7) 7) 7) 7) 7) 7) 7) 7) 7)
→ → → → → → → → → → → → → →
210k+1 210k+121 210k+151 210k+61 210k+181 210k+127 210k+37 210k+67 210k+187 210k+97 210k+169 210k+79 210k+109 210k+19
1 (mod 3) 4 (mod 5) 6 (mod 7)
→
210k+139
1 1 1 1 1 1 1 1 1 1 1 1 1 1
(mod (mod (mod (mod (mod (mod (mod (mod (mod (mod (mod (mod (mod (mod
3) 3) 3) 3) 3) 3) 3) 3) 3) 3) 3) 3) 3) 3)
1 1 1 1 1 2 2 2 2 2 4 4 4 4
(mod (mod (mod (mod (mod (mod (mod (mod (mod (mod (mod (mod (mod (mod
Denise Vella-Chemla
5) 5) 5) 5) 5) 5) 5) 5) 5) 5) 5) 5) 5) 5)
1 2 4 5 6 1 2 4 5 6 1 2 4 5
(mod (mod (mod (mod (mod (mod (mod (mod (mod (mod (mod (mod (mod (mod
Goldbach Conjecture Study
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Examples study : Example 3 Here are some examples of Goldbach decomponents belonging to arithmetic progressions found for some even numbers of the arithmetic progression 210k + 38 248 : 7 19 37 67 97 109 458 : 19 37 61 79 109 127 151 181 229 (2p) 668 : 7 37 61 67 97 127 181 211 229 271 331 878 : 19 67 109 127 139 151 271 277 307 331 337 379 421 439 (2p) 1088 : 19 37 67 79 97 151 181 211 229 277 331 337 349 379 397 457 487 541 1298 : 7 19 61 67 97 127 181 211 229 277 307 331 379 421 439 487 541 547 571 607 Conclusion : It works, of course, it is studied for. Denise Vella-Chemla
Goldbach Conjecture Study
23/5/2012
9 / 31
We try to demonstrate the impossibility that exists an even natural integer that doesn’t verify C.G. (∃ x ∈ 2N\{0, 2, 4}, x ≥ 4.1018 , x doesn0 t verify C.G.) ⇒ false but
∃ x ∈ 2N\{0, 2, 4}, x ≥ 4.1018 , x doesn0 t verify C.G.
⇔ ∃ x ∈ 2N\{0, 2, 4}, x ≥ 4.1018 , ∀ p ∈ P∗1 (x), x − p is compound ⇔ ∃ x ∈ 2N\{0, 2, 4}, x ≥ 4.1018 , ∀ p ∈ P∗1 (x), ∃ m ∈ P∗2 (x), x − p ≡ 0 (mod m) ⇔ ∃ x ∈ 2N\{0, 2, 4}, x ≥ 4.1018 , ∀ p ∈ P∗1 (x), ∃ m ∈ P∗2 (x), x ≡ p (mod m)
Denise Vella-Chemla
Goldbach Conjecture Study
23/5/2012
10 / 31
We try to demonstrate the impossibility that exists an even natural integer that doesn’t verify C.G. ∃ x ∈ 2N\{0, 2, 4}, x ≥ 4.1018 , ∀ p ∈ P∗1 (x), ∃ m ∈ P∗2 (x), x ≡ p (mod m) Quantificators expansion p1 , . . . , pk ∈ P∗1 (x), m1 , . . . , ml ∈ P∗2 (x). ∃ x ∈ 2N\{0, 2, 4}, x ≥ 4.1018 , ∀ i ∈ [1, k], ∃ j ∈ [1, l] x ≡ pi (mod mj )
Denise Vella-Chemla
Goldbach Conjecture Study
23/5/2012
11 / 31
We try to demonstrate the impossibility that exists an even natural integer that doesn’t verify C.G. Let us write all of the congruences : p1 , . . . , pk ∈ P∗1 (x), mj1 , . . . , mjk ∈ P∗2 (x). ∃ x ∈ 2N\{0, 2, 4}, x ≥ 4.1018 , x ≡ p1 (mod mj1 ) x ≡ p2 (mod mj2 ) S0 ... x ≡ pk (mod mjk )
Note : mi moduli are odd prime natural integers that are not mandatory all differents.
Denise Vella-Chemla
Goldbach Conjecture Study
23/5/2012
12 / 31
Interlude : Chinese Remainders Theorem We call arithmetic progression a set containing natural integers of the form ax + b with a ∈ N∗ , b ∈ N and x ∈ N. A congruences system not containing contradiction can be solved by the Chinese Remainders Theorem. The Chinese Remainders Theorem establishes Qk an isomorphism between Z/m1 Z × . . . × Z/mk Z and Z/ i=1 mi Z if and only if the modules mi are two by two coprime. (∀ mi ∈ N∗ , ∀ mj ∈ N∗ , (mi , mj ) = 1)
The Chinese Remainders Theorem establishes a bijection between the set of non-contradictory congruences systems and the set of arithmetic progressions.
Denise Vella-Chemla
Goldbach Conjecture Study
23/5/2012
13 / 31
Interlude : Recall of Chinese Remainders Theorem We are looking for the set of solutions of the following congruences system S : ( x ≡ r1 (mod m1 ) x ≡ r2 (mod m2 ) ... x ≡ rk (mod mk )
We set
M=
Qk
i=1
Let us calculate Let us calculate
mi . M1 = M/m1 , M2 = M/m2 , . . . , Mk = M/mk . d1 , d2 , . . . , dk such that d1 .M1 ≡ 1 (mod m1 ) d2 .M2 ≡ 1 (mod m2 ) ... dk .Mk ≡ 1 (mod mk ) k
S’s solution is x ≡ Σi=1 ri .di .Mi (mod M)
Denise Vella-Chemla
Goldbach Conjecture Study
23/5/2012
14 / 31
Interlude : Chinese Remainders Theorem Let us try to solve : x ≡ 1 (mod 3) x ≡ 3 (mod 5) x ≡ 5 (mod 7) We set M = 3.5.7 = 105. M1 = M/3 = 105/3 = 35 M2 = M/5 = 105/5 = 21 M3 = M/7 = 105/7 = 15 x
35.y1 ≡ 1 (mod 3) 21.y2 ≡ 1 (mod 5) 15.y3 ≡ 1 (mod 7)
y1 = 2 y2 = 1 y3 = 1
≡ r1 .M1 .y1 + r2 .M2 .y2 + r3 .M3 .y3 ≡ 1.35.2 + 3.21.1 + 5.15.1 = 70 + 63 + 75 = 208 = 103 (mod 105)
that are the natural integers of the sequence : 103, 208, 313, . . . i.e. from the arithmetic progression : 105k + 103 Ambiguity, Galois theory, invariant function by a roots permutation
Denise Vella-Chemla
Goldbach Conjecture Study
23/5/2012
15 / 31
Interlude : Chinese Remainders Theorem If we had to solve nearly the same congruences system, but with one congruence less : x ≡ 3 (mod 5) x ≡ 5 (mod 7) We set M 0 = 5.7 = 35. M10 = M 0 /5 = 7 M20 = M 0 /7 = 5 x
7.y10 ≡ 1 (mod 5) 5.y20 ≡ 1 (mod 7)
y10 = 3 y20 = 3
≡ r10 .M10 .y10 + r20 .M20 .y20 ≡ 3.3.7 + 5.3.5 = 63 + 75 = 138 = 33 (mod 35)
that are natural integers from the sequence : 33, 68, 103, 138, 173, 208, 243, . . . i.e. from the arithmetic progression : 35k+33 Denise Vella-Chemla
Goldbach Conjecture Study
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Interlude : Congruence relation powerfulness ≡ is an equivalence relation. a≡b c ≡d a+c ≡b+d ac ≡ bd Let us compare the resolution of the two following systems : x ≡ 3 (mod 5) x ≡ 13 (mod 5) A: B: x ≡ 5 (mod 7) x ≡ 5 (mod 7)
A : x ≡ 3.3.7 + 5.3.5 = 63 + 75 = 138 = 33 (mod 35) B : x ≡ 13.3.7 + 5.3.5 = 273 + 75 = 348 = 33 (mod 35) Denise Vella-Chemla
Goldbach Conjecture Study
23/5/2012
17 / 31
Interlude : What does the Chinese Remainders Theorem bijection ? Chinese Remainders Theorem associates to every prime modules non-contradictory congruences system an arithemtic progression. Let us call E the set of prime modules congruences systems. Let us call E 0 the set of arithmetic progressions. E sc1 sc2 sc1 ∧ sc2
→ E0 7→ pa1 7→ pa2 7→ pa1 ∩ pa2
Moreover, (sc1 ⇒ sc2 ) ⇔ (pa1 ⊂ pa2 ) . Denise Vella-Chemla
Goldbach Conjecture Study
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Interlude : Recalls An arithmetic progression being a part of N, it admits a smallest element. We will choose in the following to represent an arithmetic progression by its smallest element. E and E 0 are two given arithmetic progressions, E ⊂ E 0 ⇒ n0 ≤ n A set E provided with a partial order relation is a lattice ⇔ ∀a ∈ E , ∀b ∈ E , {a, b} admits a least upper bound and a greatest lower bound. The set of prime modules congruences systems (all modules being differents) is a lattice provided with a partial order (based on logical implication relationship (⇒)). The set of arithmetic progressions is a lattice provided with a partial order (based on set inclusion relationship (⊂)). Denise Vella-Chemla
Goldbach Conjecture Study
23/5/2012
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Interlude : Let us observe more precisely the trc bijection intervening in Chinese Remainders Theorem What are the solutions obtained by Chinese Remainders Theorem ? Z/3Z × Z/5Z → Z/15Z (0, 0) (0, 1) (0, 2) (0, 3) (0, 4) (1, 0) (1, 1) (1, 2) (1, 3) (1, 4) (2, 0) (2, 1) (2, 2) (2, 3) (2, 4)
7→ 7 → 7 → 7 → 7 → 7 → 7 → 7 → 7 → 7 → 7 → 7 → 7 → 7 → 7 →
0 6 12 3 9 10 1 7 13 4 5 11 2 8 14
In this array, line (1, 3) 7→ 13 must be read “the set of numbers that are congruent to 1 (mod 3) and to 3 (mod 5) is equal to the set of numbers that are congruent to 13 (mod 15)”. It is interesting to notice that the same line can be read “13 is congruent to 1 (mod 3) and to 3 (mod 5)” (fractality). Peano’s arithmetic axioms : let us add (1,1) recursively from (0,0) (Succ function). Denise Vella-Chemla Goldbach Conjecture Study
23/5/2012
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Interlude : Let us observe more precisely trc bijection intervening in Chinese Remainders Theorem Z/3Z × Z/5Z × Z/7Z → Z/105Z (0, 0, 0) (0, 0, 1) (0, 0, 2) (0, 0, 3) (0, 0, 4) (0, 0, 5) (0, 0, 6) (1, 0, 0) (1, 0, 1) (1, 0, 2) (1, 0, 3) (1, 0, 4) (1, 0, 5) (1, 0, 6) (2, 0, 0) (2, 0, 1) (2, 0, 2) (2, 0, 3) (2, 0, 4) (2, 0, 5) (2, 0, 6)
7→ 7→ 7→ 7→ 7→ 7→ 7→ 7 → 7→ 7→ 7→ 7→ 7→ 7→ 7 → 7→ 7→ 7→ 7→ 7→ 7→
0 15 30 45 60 75 90 70 85 100 10 25 40 55 35 50 65 80 95 5 20
(0, 1, 0) (0, 1, 1) (0, 1, 2) (0, 1, 3) (0, 1, 4) (0, 1, 5) (0, 1, 6) (1, 1, 0) (1, 1, 1) (1, 1, 2) (1, 1, 3) (1, 1, 4) (1, 1, 5) (1, 1, 6) (2, 1, 0) (2, 1, 1) (2, 1, 2) (2, 1, 3) (2, 1, 4) (2, 1, 5) (2, 1, 6)
7→ 7 → 7 → 7 → 7 → 7 → 7 → 7 → 7 → 7 → 7 → 7 → 7 → 7 → 7 → 7 → 7 → 7 → 7 → 7 → 7 →
21 36 51 66 81 96 6 91 1 16 31 46 61 76 56 71 86 101 11 26 41
(0, 2, 0) (0, 2, 1) (0, 2, 2) (0, 2, 3) (0, 2, 4) (0, 2, 5) (0, 2, 6) (1, 2, 0) (1, 2, 1) (1, 2, 2) (1, 2, 3) (1, 2, 4) (1, 2, 5) (1, 2, 6) (2, 2, 0) (2, 2, 1) (2, 2, 2) (2, 2, 3) (2, 2, 4) (2, 2, 5) (2, 2, 6)
7→ 7→ 7→ 7→ 7→ 7→ 7→ 7 → 7→ 7→ 7→ 7→ 7→ 7→ 7 → 7→ 7→ 7→ 7→ 7→ 7→
42 57 72 87 102 12 27 7 22 37 52 67 82 97 77 92 2 17 32 47 62
(0, 3, 0) (0, 3, 1) (0, 3, 2) (0, 3, 3) (0, 3, 4) (0, 3, 5) (0, 3, 6) (1, 3, 0) (1, 3, 1) (1, 3, 2) (1, 3, 3) (1, 3, 4) (1, 3, 5) (1, 3, 6) (2, 3, 0) (2, 3, 1) (2, 3, 2) (2, 3, 3) (2, 3, 4) (2, 3, 5) (2, 3, 6)
7→ 7 → 7 → 7 → 7 → 7 → 7 → 7 → 7 → 7 → 7 → 7 → 7 → 7 → 7 → 7 → 7 → 7 → 7 → 7 → 7 →
63 78 93 3 18 33 48 28 43 58 73 88 103 13 98 8 23 38 53 68 83
(0, 4, 0) (0, 4, 1) (0, 4, 2) (0, 4, 3) (0, 4, 4) (0, 4, 5) (0, 4, 6) (1, 4, 0) (1, 4, 1) (1, 4, 2) (1, 4, 3) (1, 4, 4) (1, 4, 5) (1, 4, 6) (2, 4, 0) (2, 4, 1) (2, 4, 2) (2, 4, 3) (2, 4, 4) (2, 4, 5) (2, 4, 6)
7→ 7→ 7 → 7→ 7 → 7 → 7 → 7→ 7→ 7→ 7→ 7→ 7→ 7 → 7→ 7→ 7→ 7→ 7→ 7→ 7 →
84 99 9 24 39 54 69 49 64 79 94 4 19 34 14 29 44 59 74 89 104
Same remark as for previous page concerning the two possible manners to read each line. Denise Vella-Chemla
Goldbach Conjecture Study
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restricted trc bijection Let us define restricted trc bijection as the bijection that associates to a congruences system the smallest natural integer of the arithmetic progression the Chinese Remainders Theorem associates to it. Consequence of the fact that trc (and restricted trc) are bijections restricted trc bijection associating to each prime modules congruences system with modules some mi all differents Qk a natural integer from the finite part N that is between 0 and i=1 mi , if sc1 ⇒ sc2 and sc1 6= sc2 then the sc1 congruences system solution (the natural integer paired with sc1 by restricted trc bijection) is strictly greater than the sc2 congruences system solution.
Denise Vella-Chemla
Goldbach Conjecture Study
23/5/2012
22 / 31
Let us provide an example of paired integer by restricted trc bijection of a t-uple and of the t-uples that are its projection according to some of its coordinates Let us study 3-uple (1, 4, 3) projections. Z/3Z × Z/5Z × Z/7Z (1, 4, 3)
→N 7→ 94
Z/3Z × Z/5Z (1, 4)
→N 7→ 4
Z/3Z × Z/7Z (1, 3)
→N 7→ 10
Z/5Z × Z/7Z (4, 3)
→N 7→ 24
94 has three integers paired with itself that are strictly lesser than it by restricted trc bijection. 94 projects itself in natural integers strictly lesser than it because 3.5 < 3.7 < 5.7 < 94 < 3.5.7. Denise Vella-Chemla
Goldbach Conjecture Study
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Interlude : Fermat’s Infinite Descent If an even integer should not verify Goldbach Conjecture, there should be another even integer lesser than the first one that should not verify Goldbach Conjecture neither and step by step, like this, we proceed until reaching so little integers that we know they verify Goldbach Conjecture. Exists no infinite strictly decreasing sequence of natural integers. Reductio ad absurdum : - we suppose that x is the smallest such that P(x). - we show that then P(x 0 ) with x 0 < x. - we reached a contradiction. (If P(n) for a given natural integer n, there exists a non-empty part of N containing an element that verifies property P. This part admits a smallest element. In our case, property P consists in not verifying Goldbach Conjecture)
Denise Vella-Chemla
Goldbach Conjecture Study
23/5/2012
24 / 31
Recall : We try to reach a contradiction from the following hypothesis : p1 , . . . , pk ∈ P∗1 (x), mj1 , . . . , mjk ∈ P∗2 (x). ∃ x ∈ 2N\{0, 2, 4}, x ≥ 4.1018 , x ≡ p1 (mod mj1 ) x ≡ p2 (mod mj2 ) S0 ... x ≡ pk (mod mjk )
Note : some modules can be equal.
Denise Vella-Chemla
Goldbach Conjecture Study
23/5/2012
25 / 31
First step System transformation to order modules according to an increasing order and to eliminate redundancies. p10 , . . . , pk0 ∈ P∗1 (x), nj1 , . . . , njk ∈ P∗2 (x). ∃ x ∈ 2N\{0, 2, 4}, x ≥ 4.1018 , x ≡ p10 (mod nj1 ) x ≡ p20 (mod nj2 ) S ... x ≡ pk0 (mod njk ) S has d that is paired with itself by restricted trc bijection.
Denise Vella-Chemla
Goldbach Conjecture Study
23/5/2012
26 / 31
Where can contradiction come from ?
It can come from Fermat’s Infinite Descent. We know that restricted trc bijection provides as solution for S the natural integer d that is the smallest natural integer of the arithmetic progression associated to S by the Chinese Remainders Theorem. S system is such that d doesn’t verify Goldbach Conjecture. Conclusion : We are looking for a S 0 congruences system, implied by S and 6= to S, to what restricted trc bijection associates a natural integer d 0 < d, with d 0 doesn’t verify Goldbach Conjecture neither.
Denise Vella-Chemla
Goldbach Conjecture Study
23/5/2012
27 / 31
We look for S 0 ⇐ S that has d 0 < d paired with it by restricted trc bijection. Let us consider a S 0 congruences system constituted by a certain number of congruences of S according to all differents odd prime natural Q integers mi , i an integer between 1 and k, such that d > ki=1 mi ; First problem : To descend one Fermat’s descent step, it is necessary that d 0 < d. But we saw that d 0 < d comes from restricted trc bijection. Second problem : How to be sure that d 0 doesn’t verify Goldbach Conjecture neither ? For this aim, we need that congruences kept from initial S congruences system are such that d 0 is congruent to all P∗1 (d 0 ) elements according to a module that is an element of P∗2 (d 0 ). (Said in another way, we must be sure that removing congruences to make strictly decrease the congruences system solution, we won’t “lose” congruences that ensured the Goldbach Conjecture non-verification.) Denise Vella-Chemla
Goldbach Conjecture Study
23/5/2012
28 / 31
Second step We keep from the resulting congruences system a maximum of congruences in a new system S 0 in such a way that d, the initial system S’s solution, is strictly greater than the moduli kept in the new system product and in such a way that every√module intervening in a kept congruence of the system is lesser than d 0 . . p10 , . . . , pk0 0 ∈ P∗1 (x), nj1 , . . . , njk 0 ∈ P∗2 (x). ∃ x ∈ 2N\{0, 2, 4}, x ≥ 4.1018 , 0 x ≡ p10 (mod nj1 ) x ≡ p2 (mod nj2 ) S0 ... x ≡ pk0 0 (mod njk 0 ) Q 0 d > ku=1 nju px0 are all differents odd prime natural integers and ny are all differents odd prime natural integers ordered according to an increasing order. S 0 is paired with d 0 by restricted trc bijection. Denise Vella-Chemla
Goldbach Conjecture Study
23/5/2012
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Why d 0 doesn’t verify Goldbach Conjecture neither ? d0
